Problem: Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic. 

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

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The dragon has a hit point of x initially. When its hit point goes to 0 or under 0, it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells. 

  * Void Absorption

Assume that the dragon's current hit point is h, after casting this spell its hit point will become \left⌊ h/2 \right⌋ + 10. Here \left⌊ h/2 \right⌋ denotes h divided by two, rounded down.

  * Lightning Strike

This spell will decrease the dragon's hit point by 10. Assume that the dragon's current hit point is h, after casting this spell its hit point will be lowered to h-10.




Due to some reasons Kana can only cast no more than n Void Absorptions and m Lightning Strikes. She can cast the spells in any order and doesn't have to cast all the spells. Kana isn't good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next t lines describe test cases. For each test case the only line contains three integers x, n, m (1≤ x ≤ 10^5, 0≤ n,m≤30) — the dragon's intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Example

Input


7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2


Output


YES
NO
NO
YES
YES
YES
YES

Note

One possible casting sequence of the first test case is shown below:

  * Void Absorption \left⌊ 100/2 \right⌋ + 10=60.
  * Lightning Strike 60-10=50.
  * Void Absorption \left⌊ 50/2 \right⌋ + 10=35.
  * Void Absorption \left⌊ 35/2 \right⌋ + 10=27.
  * Lightning Strike 27-10=17.
  * Lightning Strike 17-10=7.
  * Lightning Strike 7-10=-3.

Solution (in PYTHON3):

Solution:
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from functools import lru_cache
import bisect
import re
import queue
from decimal import *


class Scanner():
    @staticmethod
    def int():
        return int(sys.stdin.readline().rstrip())

    @staticmethod
    def string():
        return sys.stdin.readline().rstrip()

    @staticmethod
    def map_int():
        return [int(x) for x in Scanner.string().split()]

    @staticmethod
    def string_list(n):
        return [input() for i in range(n)]

    @staticmethod
    def int_list_list(n):
        return [Scanner.map_int() for i in range(n)]

    @staticmethod
    def int_cols_list(n):
        return [int(input()) for i in range(n)]


class Math():
    @staticmethod
    def gcd(a, b):
        if b == 0:
            return a
        return Math.gcd(b, a % b)

    @staticmethod
    def lcm(a, b):
        return (a * b) // Math.gcd(a, b)

    @staticmethod
    def divisor(n):
        res = []
        i = 1
        for i in range(1, int(n ** 0.5) + 1):
            if n % i == 0:
                res.append(i)
                if i != n // i:
                    res.append(n // i)
        return res

    @staticmethod
    def round_up(a, b):
        return -(-a // b)

    @staticmethod
    def is_prime(n):
        if n < 2:
            return False
        if n == 2:
            return True
        if n % 2 == 0:
            return False
        d = int(n ** 0.5) + 1
        for i in range(3, d + 1, 2):
            if n % i == 0:
                return False
        return True


def pop_count(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f


MOD = int(1e09) + 7
INF = int(1e15)


def solve():
    X, N, M = Scanner.map_int()
    for _ in range(N):
        Y = X // 2 + 10
        if Y < X:
            X = Y
        else:
            break
    X -= 10 * M
    print('YNEOS'[not X <= 0::2])


def main():
    # sys.stdin = open("sample.txt")
    T = Scanner.int()
    for _ in range(T):
        solve()


if __name__ == "__main__":
    main()